Big try at gravity wheel

[TinselKoala]

You have to put it in, somehow, before you can take it out. And unlike the USA economy, you can't take out more than you put in, in the first place. That's the rub.   

[MarkE]

I understand that,, but the floatability of the float with a fixed shape does not care if it is 5 miles down or 5 inches down, the buoyancy is the same amount.

This then turns the float into a constant force and that force does not care about a change in height.

To move the float in the opposite direction that it prefers to move takes an input,, but to let it move where it wants does not, as a mater of fact, as you even stated, that is work that can be taken out.

Webby, the buoyant force is the weight of the displaced water which doesn't change much even 5 miles down, we agree.  The equations that describe the work going up or going down are independent of the float's weight.  What matters is that the net work performed in a closed cycle from one depth over any path back to that same starting depth is zero.  You can do whatever you want, but by the time you finish a cycle and return to the starting point for the next cycle, ignoring losses, the work available, and the work performed are both zero.  That's true independent of the "float's" SG.  The float could be filled with air or lead balloons and the mechanics are the same.  SG only determines whether work has to be applied to move up or to move deeper down.

One can no more gain energy with a buoyancy machine of any kind than one can gain energy with a coil spring.  In a buoyancy machine, the system can release energy once:  for SG > 1 going down, and for SG < 1 going up.  In order to return to the starting position, the same amount of energy has to be returned as the system gave up.

[LibreEnergia]

I understand that,, but the floatability of the float with a fixed shape does not care if it is 5 miles down or 5 inches down, the buoyancy is the same amount.

This then turns the float into a constant force and that force does not care about a change in height.

To move the float in the opposite direction that it prefers to move takes an input,, but to let it move where it wants does not, as a mater of fact, as you even stated, that is work that can be taken out.

But it hasn't occurred to you that no NET work is produced when cycling the float through any range of depths and ending up at the starting location?

As and aside, do you think it would take more work, or the same to raise the Titanic  from its current depth of 3800m compared to a depth of say 100 metres.

[MarkE]

With no changes I agree, with most changes they have a cost, I agree.

*IF* this system can effect a change at a lesser cost by changing the environmental relationships then things are not what is normal.  If this system can change those values so that a larger portion of the input work can be recovered from the system then the savings in that recovery are the gain.

No one at HER, nor any of their supporters has shown ANY evidence that: "this system can effect a change at a lesser cost".  Nor have they shown that any system can.  The statement itself is circular:  "If I could get a free lunch, I would eat for free."  The fact remains that carrying a mass up and down from a starting point back to the same starting point whether in:  a vacuum, or a fluid is in the best idealized case conservative.  In all real cases it is lossy.

Yes it has TK.

In the most basic view this system changes volume for pressure,, that is more input pressure but at less volume.  The system has no issue with having the lift be 100 percent efficient,, unless you are willing to say that energy can be destroyed, in reality I would doubt a 100 percent efficient lift when the input by itself is considered,, but myself,, I had way in excess of 15 percent, that is what was predicted IIRC.

Lift is a force.  Force is not conservative.  Anyone who has ever seen a lever in action knows that force is not conservative.  Energy on the other hand is conservative.  One can use any of many devices to manipulate force, but none of those measures alone or in any combination will gain output energy over the input energy.

*IF* after the lift, the stored potential is used to initiate a second cycle, then is it not probable that that input will reduce the actual input value needed by what ever value of stored potential is left?

Read again:  At the end of any cycle that returns all elements to the original starting point there is no additional energy anywhere at the end of the cycle than at the start.

At the end of lift the full pressure is still within the closed system, that pressure is above the pressure of a second system at the rest pressure,, which way will the fluid and pressure move if the two are connected together,, have you not considered that the pressure and therefore fluid MUST move from the higher value to the lower one?

Pressure like force is not conservative.  Energy:  pressure times volume is conservative.

*IF* I have my reservoir refilled 1\2 way when it has dropped 1\2 the distance,, is in not fair to say that I have a fair amount of potential coming back out from the system?

Should the actual question be: How much potential can be recovered from the system after lift?

The actual question should be:  How much energy can be removed in a cycle and complete that cycle with the same energy as at the cycle start?  The best case answer is zero.  The answer in any real system with losses is negative.  Net energy has to be supplied in order to sustain one cycle to the next or else the system runs down and stops.  Which is exactly what happens to the ZED and the TAZ.

You and I know that the pressure is still at its full value at the end of lift, then the question is in the rate of decrease in the pressure as the fluid is removed, and since the weight left on the system is the required weight to hold the system at rest pressure that value is the base line, that value is the lowest pressure value the system will fall to.

As TinselKoala's demonstration shows, the stored potential energy runs down.  Pressure alone does not define energy.  Pressure times volume defines energy.

I had many lifts that were 75 percent or better, so how much of that stored potential is needed for recovery,, that would be 25 percent,, correct?  Any more than 25 percent with a 75 percent efficient lift would be extra. 

In other words you never broke even.  Neither have HER broken even.  Nor will HER ever break even with energy.

My values were for the removed weight after lift to water weight to make the lift and the distances they both moved,, you know that but I thought I would put that back out there.

That then starts with the straight question of the expanding air, how much volume increase happens when the air expands to 1\3 its pressure, this is not even taking the funny behavior into account of the riser response difference.

If this is enough to get the second system from rest to lift potential then your Bollard effect happens.

[LibreEnergia]

*IF* after the lift, the stored potential is used to initiate a second cycle, then is it not probable that that input will reduce the actual input value needed by what ever value of stored potential is left?

...if you have an endless supply of floats at the bottom of your tank, then yes you could. However most normal tanks are finite and only contain a few floats. Once they rise to the surface then the ability to tap that potential from them is gone.  To reset you need to get the floats to the bottom of the tank again.

No matter WHAT way you try to do that, there is an input cost exactly equal (or more) than sinking the float to the bottom. Even if you drain the tank, drop the float and refill it that energy cost does not go away. (In this case the energy input is in pumping the water up).

The system described by Travis just does not work.